 Now, let us move on to the next lecture on surface potential and threshold based solutions of the drain current for a large uniformly doped bulk MOSFET under DC bias, a quick recap. What we are doing in the previous lecture is that we were working out a solution for psi s as a function of gate to bulk voltage and channel voltage from y dimensional analysis. So, the outline of the solution was that using gradual channel approximation on the left hand side and the fact that Jn y, Jpy is equal to 0 approximately which is another way of stating the Boltzmann relation on the right hand side of the Poisson's equation. We get the y dimensional Poisson-Boltzmann relation written here where Kpp0 and Np0 into exponential of minus V by Vt are values at the edge of the space charge layer. And we said that if you integrate this equation and approximate it, we can show that in inversion the Qs as a function of psi s that is the total silicon charge as a function of psi s can be given by this formula okay where Ld is the Debye length. Now, let us quickly understand the terms of this equation once again and continue with our derivation. So here let us start with the left hand side of this equation dou square psi by dou y square the psi. So this is the variation of psi as a function of y so we are looking at the MOSFET at some x here in the y direction this is the depletion width and potential psi is a potential at any point here with respect to the depletion edge. So this potential is psi s and the potential of any point here with respect to yd will be psi the same thing is sketched here. So this is psi as a function of y the psi goes on decreasing to 0 this is what is shown here. So psi is any potential is a potential psi at a psi is a potential at any point y okay. Now let us move on to the right hand side of the equation the k times pp0 what is k times pp0? So k times pp0 is the whole concentration at the depletion edge okay. Then np0, np0 is the equilibrium concentration of electrons at the depletion edge that is what is shown here okay this is np0 here and what is np0 into exponential of minus v by vt okay that is the concentration at the depletion edge for electrons so this is that concentration okay. Now what is kpp0 into exponential of minus psi by vt this is the concentration of holes at any point so if I were to take any y now this is p the whole concentration and this p is nothing but this quantity you can see that it will be written as the boundary condition of the whole concentration at the depletion edge multiplied by exponential of minus psi by vt you know this whole concentration is decreasing so if you want to use Boltzmann relation you want to relate this concentration of holes to the potential psi here okay at the same point. Now similarly if you take this term np0 into exponential of minus v by vt into exponential of psi by vt that is the electron concentration here n now this is obtained by multiplying the concentration of electrons at the depletion edge by exponential of psi by vt where psi is the potential at the same point here and what is v so v is the channel voltage so here when we write the concentration of electrons at the depletion edge it is written in terms of the voltage applied between the induced n plus p junction and the channel voltage is the voltage at this point. So please distinguish between v and psi s psi s is the total potential drop across the depletion layer okay whereas v is the channel voltage only part of the voltage that is applied and that is created by the vdb applied at the drain and vsb applied at the source and it also depends on the vgb that is applied to the gate so I have not shown the gate here so you have the gate and here you are applying gate voltage okay and nA in this equation is the doping of this p region so I hope that now all the terms of the equations are clear now let us move on to the derivation okay so how do we derive this equation for qs from the Poisson Boltzmann relation so this was the equation we had written in the previous lecture this equation in a normalized form okay the Poisson Boltzmann relation or Poisson Boltzmann equation is written in a normalized form where here psi star is psi by vt then phi f star is phi f by vt and v star is equal to v by vt the term y star is equal to y by the debilength ld where the ld is given by square root epsilon is vt by q times the majority carrier concentration in the substrate now how do we integrate this equation so what we do is we multiply both sides by the term 2 psi star by dy star okay now what is the idea in doing this the idea is when you multiply left hand side by this term you can recognize the left hand side as the differential of dou psi star by dou y whole square okay so let us do that so LHS is equal to into dou psi star by dou y star into dou square psi star by dou y star square and this is nothing but dou by dou y star of dou psi star by dou y star whole square so this is what will be left hand side now here we know that dou psi star by dou y star is nothing but negative of the electric field in the y direction in the normalized form okay so left hand side therefore becomes equal to so therefore LHS is equal to dou by dou y star of E y star square because though there is a negative sign here since we are squaring it after squaring the negative sign goes away okay so when we multiply the left hand side by this term the left hand side is reducing to dou by dou y star of E y star square now so what we will do is from both sides we will remove the term dou y star so remove dou y star from both sides that is after multiplying the right hand side by this okay so when you remove that what will be left with is only dou by E y star square and your equation will become dou of E y star square is equal to minus 2 into this whole term here that is k into exponential of minus y star minus 1 minus exponential of minus twice 5 star the exponential of psi star minus v star minus 1 dou psi star okay so you can see that right hand side is a function of y star alone and that is being multiplied by dou y star so now it will be easy to integrate both sides okay because when I integrate dou of E y star square I will get simply E y star square right and right hand side the exponential terms and so on can be integrated very easily so the result of this integration would be as follows so integrate from y star equal to bulk y star equal to 0 right so you are going in the direction from bulk to the surface y star equal to 0 is the surface so y star equal to bulk is E y star equal to 0 whereas y star equal to 0 E y star is equal to E y s star right that is the surface value of the electric field at y star equal to 0 and corresponding values of psi star in the bulk psi star is equal to 0 and at the surface y star equal to 0 that is surface you have psi star equal to psi s star so now the result of this integration would be the following now let me rub off this part so result of integration and putting the limits I will get E y s star square is equal to 2 times psi s star minus k into 1 minus E power minus psi s star plus 2 times exponential of minus phi s star into exponential of minus V star into exponential of psi s star minus 1 minus psi s star okay so let us look at this so left hand side you have got this square term itself after integration right hand side you can see that this term 1 when you integrate okay you will get psi s star minus 0 when you integrate 1 you will get psi s star and when you put the limits of psi star that is 0 and psi s star you will end up getting the psi s star term so that is how you are getting this term then exponential of minus psi s star when you integrate you will get negative of minus psi s star and that you integrate from psi s you will get negative of minus psi star that you integrate from again psi star equal to 0 you put the limits psi star equal to 0 and psi s star and you will end up getting k times exponential of minus psi s star 1 minus exponential of minus psi s star right similarly go to the next equation here exponential of minus V star is the constant when you are integrating with respect to psi star that is why this is brought out and the exponential of psi star when you integrate between the limits psi star equal to 0 and psi star equal to psi s star you will get exponential of psi s star minus 1 and finally this term 1 again when you integrate you will get psi s star right like you got psi star here okay so now whenever we get a complicated equation like this we must check whether we have taken care of all the terms right it is very important to have a care for the details whenever we are manipulating equations particularly lengthy equations in the context of modeling of semiconductor devices. So one way to check this is check whether E y s star goes to 0 for psi s star tending to 0 okay let us check this so if you put psi star equal to 0 this term becomes 0 here you put 0 exponential of 0 is 1 so this term goes to 0 then here again you put psi star equal to 0 this term goes to 0 and this term also goes to 0 so this limiting condition is satisfied okay that is one way of checking this long equation. Now let us write this equation for E y s star right so you will get a square root on the right hand side so you have the equation E y s star is equal to so when you take a square root you can get both plus and minus so plus and minus so instead of 2 you will get square root of 2 into square root of this whole term which I am writing again minus k into 1 minus the minus psi star plus okay I think this part I will slightly simplify it and write it in this form okay so you multiply E power psi star by E power minus V star you get this and then you get exponential of minus V star that is taken here and this psi star is put here okay so this is your equation for E y s star now what is E y s star it is a normalized electric field so let me write here what is E y s star so E y s star is equal to E y s divided by the normalized divided by V t by L d so you can see that V t by L d has the unit of Debye length sorry V t by L d has the unit of electric field okay so you are normalizing the dimension Y with respect to L d here okay so that is how when you want to normalize the electric field you will have to normalize it with respect to V t by L d now let us try to see where will you use the positive sign and where will you use the negative sign so the positive sign is to be used when psi s star is more than 0 that is in depletion or inversion and negative sign is used when psi s star is less than 0 that is in accumulation so here is an assignment for you using the expression for E y s star sketch Q s versus psi s on a linear and similar scales for V star equal to 0 so in that expression for E y s star you substitute V star equal to 0 okay and then you plot this expression now I want to remind you that the value of Q s is related to E y s as follows so Q s is nothing but epsilon s into E y s okay now let us consider some approximations of this long E y s star equation okay now we are going to operate in depletion or inversion and not in accumulation so we are going to use only the positive sign so let me rub this part off now how do you approximate so let us look at some numbers in inversion where we are going to use this equation extensively you have psi s star more than V star plus phi f phi f star right so when you take a two term you take a two terminal MOS capacitor now you recall that once the surface potential exceeds phi f then your concentration of electrons at the surface exceeds ni okay and then you start inversion because at the surface your electron concentration is more than whole concentration okay so when you have the channel voltage V then the condition for inversion on the surface potential would be psi star is more than V star plus phi f star now what is the value of phi f star in typical cases so phi f star lies between 13 and 18 okay so phi f star is phi f by V t and you know what is phi f by V t so phi f is equal to V t ln na by ni okay that is how it depends on doping so you have to calculate this and phi f by V t you remove the V t from here so this is just ln of na by ni so what is your range of na so range of na is 10 power 16 to 10 power 18 per centimeter cube okay so for this range of na this quantity phi f by V t will vary from 13 to 18 so this term is really 13 to 18 okay and your psi star therefore will be more than 13 to 18 because you are adding V star so in fact even if phi f star is a few V t okay this means that you can be in depletion even then the following approximation will hold right that is your psi star will be much more than k so that will allow you to approximate things here in this and then e power minus psi star will be much less than 1 that will help you to approximate this term and exponential of psi star minus V star will be much more than psi star minus exponential of minus V star so here you have psi star minus exponential of minus V star right so this will help you to neglect this term let us make this approximations so if you expand this bracket you will get minus k minus a plus k into exponential of minus psi star so since psi star is much more than k and exponential of minus psi star is much less than 1 so this is much less than 1 so I can neglect this and psi star is much more than k so I can neglect this whole term and then coming here since exponential of psi star minus V star so psi star minus V star is phi f star here so which lies between 13 and 18 so exponential of 13 to 18 that is really a very large number compared to psi star itself okay and V star is more than 1 and therefore exponential of minus V star so what is V star? V star is normalized channel voltage V by V t so this normalized quantity will be more than 1 and therefore this quantity will be very small okay so in any case this is subtracting from psi star so that will reduce the right hand side quantity okay so the left hand side is much more than this in other words I can simply cancel out these 2 terms so what I therefore get is a much simpler expression as follows so E of Y S star is equal to I remove the negative sign because I am not going to use it in accumulation square root 2 into square root of psi star plus I think I made a small mistake here these 2 was not necessary okay so that 2 should be removed so psi f star plus exponential of psi f star minus now twice phi f star and V star we can sum up and club together and put it as V star plus phi f star club together okay and this is approximately equal to and as a result this leads to the formula for Q S so Q S is equal to epsilon S E Y S and that you can write in terms of E Y S star is equal to E Y S star into V t by L D okay so because E Y S star is E Y S by V t by L D so E Y S is E Y S star into V t by L D okay and so that can be written as now let me write it here I am sorry Q S so the formula for Q S is epsilon S V t by L D into root 2 into square root of psi star that is I am taking from here plus exponential of psi star minus V star plus twice phi f star okay now whenever we write a charge or field we must be careful about the sign right so now when we wrote the charge here for example we know that we are considering an n type MOSFET so the silicon charge is negative because it is due to electrons and ionized donors therefore actually I should put a negative sign here okay so the surface electric field E Y S is in the positive y direction sorry y direction is from top to bottom like this from gate to bulk and the electric field is also from gate to bulk in this direction because gate is positive with respect to bulk so that is why E Y is in the positive y direction therefore E Y is positive but when E Y is positive your charge is negative right Q S is a silicon charge that is negative so there has to be a negative sign here so I should put a negative sign here and I should put a negative sign here okay now I leave it to you to show that the same equation can also be written in another form and that is minus of C ox into gamma into square root of this whole term comes here okay so the coefficient here can be written as minus of C ox into gamma so that is in fact the equation for Q S as a function of psi S written here okay where the psi S star has been written in the unnormalized form right so that is psi S by V T and similarly V star is V by V T and twice phi F star is twice phi F by V T okay so the same equation is written here is the equation for Q S okay now ultimately we need to go to the psi S so that is one more step okay let us see how we do that so now substituting Q of psi S Q S of psi S into surface potential equation with psi ox equal to minus of Q S plus Q F by C ox which is expressed using the gradual channel approximation right because this is like a capacitor formula where the conditions are one dimensional but in MOSFET it is two dimensional right so we said that we can neglect the variation of the field in the x direction because that is slower than the variation in the y direction and make it approximately one dimensional from gate to bulk side and write this so using a gradual channel approximation in Gauss law so we get in inversion this relation okay where the value of psi S is psi S 0 at the source or for channel voltage V equal to VSB so if you put V equal to VSB you will get the value of psi S at the source so this will be psi S 0 and this will also be psi S 0 this is an implicit equation for psi S okay and similarly you can put channel voltage V equal to VDB and then you will get psi SL so these terms will be psi SL. Now let us show how we can get this from this formula for Q S so the surface potential equation says that VGB is equal to psi ox plus psi S plus phi MS we have neglected the potential drop in poly psi P. Now what we are saying here is you replace psi ox by this term okay so if you replace that so using gradual channel approximation this term we are writing as minus of Q S plus Q F by C ox okay then the minus Q F by C ox term can be clubbed with phi MS and we know that phi MS minus Q F by C ox is flat band voltage so that is how you are getting the flat band voltage here okay and then the Q S is written in terms of so this Q S is minus gamma C ox into square root of all these terms right which we have seen here so all these things here we will put it in this form actually so minus C ox gamma into square root of all of this. So when you multiply when you divide this Q S by C ox and take the negative sign off then this C ox will get removed and that is how here you do not have the C ox term okay so this is a straight forward thing I leave the remaining simple manipulations to you okay and show that just from this equation you can get this equation for psi S this is a so called surface potential equation okay now here is an assignment for you sketch psi S as a function of channel voltage V and mark the depletion weak inversion moderate inversion and strong inversion regions of operation so let me give you the hint how that function is going to look like so this is your V axis and this is your psi S axis now when your channel voltage is 0 here you solve this equation you will get some nonzero psi S right because you do have the depletion inversion layers and then as you increase your V this surface potentially will go on increasing because you know that when you move from source to drain your surface potential goes on increasing right so this is your picture and this is your surface potential it goes on increasing channel voltage also goes on increasing so as you increase your V you will find that your surface potential increase and ultimately it will tend to saturate okay this is what you have to show for large V your surface potential will not change much and in this region you have to identify the various regimes right such as depletion weak inversion moderate inversion and strong inversion right so this I will leave it to you which of this region is weak inversion which of this region is strong inversion okay which of this region is moderate inversion which is depletion right one more assignment using the expressions for E Y S and psi S obtained from the solution of Poisson-Boltzmann equation calculate the E Y S and psi S at the surface of a two terminal NMOS structure having the following parameters N plus polysilicon gate the oxide thickness T ox is 10 angstroms the fixed charge is 5 into 10 power 10 per centimeter square the substrate doping NA is 5 into 10 power 17 per centimeter cube and the gate to bulk biac is 2 volts temperature is 300 K find out by what percentage does the calculated psi S differ from the commonly assumed value of twice phi F in strong inversion so now let us put all our equations together okay as a summary so the surface potential based model for drain current as a function of gate to bulk drain to bulk and source to bulk voltage can be written as follows so the drain to source current consists of two components the diffusion component and drift component the diffusion component is written in this form where psi S L is a surface potential at the drain and psi S naught is a surface potential at the source and this is the average surface mobility okay then the drift current is written in this form again in terms of the surface potential at the drain and surface potential at the source where the surface potential is calculated using this formula so you get the surface potential at the source when you put the channel voltage V here as VSB the source to bulk voltage and you get the surface potential at the drain when you put the channel voltage as VDB or drain to bulk voltage now what is important is this that above psi S expression saturates with increase in channel voltage hence this model predicts the ideas behavior over the entire bias range without the need for any partitions such as sub or super threshold non saturation and saturation and weak and strong inversion so this is a model it is not a piecewise model it is one equation which predicts all the various regimes of operation so that is the nice thing about this model okay the only difficulty in this model is that this surface potential equation for psi S is an implicit equation and a lot of computation is involved in solving this equation right for any channel voltage okay or any source to bulk and drain to bulk voltage. So that is why in fact because of this computational load people have resorted to other forms of equations such as the threshold voltage based equations but modern MOSFETs have become so complex and so small geometry that the threshold voltage based equation also involves lot of computation and therefore people are having a relook at the surface potential based model which appears fairly simple except for this implicit equation present there and they are trying to do research on how to solve this equation in an efficient form okay in as lesser time as possible in using as much less computation as possible so lot of research has been directed in MOS modeling at solving this particular implicit equation okay efficiently.